Solve for v: "How to Solve for v with Theta, g, and x

  • Thread starter Thread starter marthkiki
  • Start date Start date
AI Thread Summary
To solve for v in the equation involving theta, g, and x, the user has derived v^2 = (v^4 - 86612.49)^(1/2) + 4.165. The next steps involve rewriting the equation to isolate terms and squaring both sides, which may lead to false solutions. The discussion highlights the importance of correctly squaring the left-hand side to form a quadratic equation in v. Users are advised to check their solutions carefully, as squaring can introduce extraneous roots. The conversation concludes with the user acknowledging the mistake in their previous calculations.
marthkiki
Messages
13
Reaction score
0

Homework Statement


theta = tan^-1 {{v^2 - [v^4-g(gx^2)]}^1/2}/gx
I know that x is 30, theta is 45, g is 9.81 m/sec^2.
I'm trying to solve for v.

Homework Equations


The Attempt at a Solution


I've gotten all the way down to
v^2 = (v^4-86612.49)^1/2 + 4.165
What do I do from here?
 
Last edited:
Physics news on Phys.org
marthkiki said:

The Attempt at a Solution


I've gotten all the way down to
v^2 = (v^4-86612.49)^1/2 + 4.165
What do I do from here?

you'll need to re-write it as


v2 - 4.165= (v4-86612.49)1/2

then square both sides and solve. You may need to use a numerical method. You'll need to test your solutions as squaring both sides might give you false answers.
 
Once you square it you will be left with only 1 V term. It may be easier to replace your constants with variables. Then just substitute them in once you find V.
 
How "only 1 V term"? There will be v^4 and v^2 terms but then you can let u= v^2 and will have a quadratic to solve for u.
 
rock.freak667 said:
you'll need to re-write it asv2 - 4.165= (v4-86612.49)1/2

then square both sides and solve. You may need to use a numerical method. You'll need to test your solutions as squaring both sides might give you false answers.

By squaring both sides, I get v^2 - 17.35 = v^4 - 86612.49.
That gets me -17.35 = -86612.49.
Is this what you mean by a false answer?
 
HallsofIvy said:
How "only 1 V term"? There will be v^4 and v^2 terms but then you can let u= v^2 and will have a quadratic to solve for u.

If i do that, I end up with u = v2 + 416.2
then the discriminant will end up negative leaving me with an imaginary velocity. Did I do this correctly?
 
If i do that, I end up with x = 86612.49 and y = 4.165. I then end up with v^4 - y^2 = v^4 - x, which again gets me to -y^2=x.
 
v^2 -4.165 = (v^4-86612.49)^1/2
v^2 - x = (v^4-y)^{1/2}
v^4 - 2v^2x + x^2=v^4-y
-2v^2x + x^2=-y
v=\sqrt{\frac{x^2+y}{2x}}
there u go
 
marthkiki said:
By squaring both sides, I get v^2 - 17.35 = v^4 - 86612.49.
The left-hand side is not squared properly. The left hand side should be (v^2 - 17.35)^2 = v^4 - 2(17.35)v^2 + (17.35)^2. The v4's will then cancel and you will be left with a quadratic in v. You will need to check the two answers you get, as squaring both sides of an equation introduces new solutions that may not solve the original equation. Ie., x = -1 and x2 = 1, which introduces the solution x = 1.
 
  • #10
Thank you so much!
I get it now. I totally squared the left wrong.
 
Back
Top